Gibbs Free Energy & Spontaneity

Pick a reaction, or type its ΔH and ΔS, and set a temperature. The page works out TΔS, ΔG and the equilibrium constant, the temperature where ΔG changes sign, and which of the four sign cases the reaction is in.

Worked example — Gibbs Free Energy & Spontaneity

The entropy term	+47.76 kJ/mol
Gibbs free energy change	+131.04 kJ/mol
ΔG for one factor of ten in K	5.708 kJ/mol
Equilibrium constant, as a power of ten	-22.96
Equilibrium constant	1.10×10⁻²³
Where ΔG changes sign	843.1 °C

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Gibbs Free Energy & Spontaneity

The Gibbs free energy change of a reaction, ΔG = ΔH − TΔS, is the most work it can deliver at constant temperature and pressure, other than pushing back the atmosphere. A reaction with ΔG < 0 can proceed by itself; one with ΔG > 0 goes the other way; at ΔG = 0 the two directions balance. The standard value fixes the equilibrium constant: ΔG° = −RT ln K. When ΔH and ΔS have the same sign, ΔG changes sign at T = ΔH/ΔS.

What each symbol means

ΔG: the Gibbs free energy change, in kJ per mole of reaction as written. Negative: the products are favoured
ΔS: the entropy change, in J/(mol·K). Its sign is the slope of ΔG against T, with the sign reversed
T: the absolute temperature, in kelvin. Every formula on this page uses it; the Celsius box is for convenience
K: the equilibrium constant, dimensionless: pressures in bar and concentrations in mol/L, each divided by its standard value

Assumes: ΔH and ΔS keep their 25 °C values at every temperature. This is the Ellingham approximation, and it is what makes the first panel a straight line. All species are in their standard states, 1 bar for gases, so ΔG here is ΔG° and K is the value that goes with it; a real mixture has a reaction quotient Q, and the reaction runs towards K from wherever Q starts. The ice preset is built differently: its ΔS is ΔH divided by the melting point, which is how an entropy of fusion is measured, so its crossover sits at 0 °C by construction.
Breaks when: At high temperature, by an amount the worked problem measures: for limestone the constant-value crossing is 45 degrees too low. And ΔG says nothing about speed. Diamond turning into graphite has ΔG ≈ −2.9 kJ/mol at room temperature, and diamonds last for billions of years, because no pathway at that temperature is fast enough to notice. The Haber process runs at 450 °C, against its own equilibrium, for the same reason in reverse.

Only the sign of ΔS decides whether heat can reverse a reaction 🖖

ΔG = ΔH − TΔS is a straight line in T, and the first panel draws it. Extrapolated back to absolute zero its height is ΔH, and its slope is −ΔS. Heating a reaction can only walk along that line, so if ΔS is positive the line falls and a reaction that refuses at room temperature eventually goes; if ΔS is negative it rises and the opposite happens. For limestone the line crosses zero at 843.1 °C, where 160.18 J/(mol·K) times the temperature finally pays for 178.8 kJ. Real kilns need about 898 °C before the carbon dioxide reaches one atmosphere, measured by Smyth and Adams in 1923, and the worked problem below finds where most of the 55 degrees went.

5.708 kJ/mol buys one factor of ten 🖖

At 25 °C, ΔG = −RT ln K becomes one line of arithmetic: every 5.708 kJ/mol of ΔG moves K by a factor of ten. Ammonia's −32.81 kJ/mol is 5.75 of those steps, so K is 5.61×10⁵. A reaction with ΔG of −5.7 kJ/mol ends up only ten to one in favour of its products, and it takes about −57 before the reactants are gone to ten decimal places. The number is not new to this site. It is nernst-cell-voltage's 59.16 mV per decade multiplied by Faraday's constant, the same fact in the currency of a battery. It also grows with temperature: at 450 °C one factor of ten costs 13.845 kJ/mol.

It absorbs heat and happens anyway 🖖

Ice needs 6.01 kJ/mol to melt, and at 10 °C it melts, because TΔS has grown past ΔH: ΔG is −0.22 kJ/mol. An endothermic reaction that goes is not a paradox; it is the top-right cell of the sign table. The bottom-right cell is the one to read carefully. Oxygen to ozone has ΔH = +285.34 and ΔS = −137.62, so no temperature makes it go, and yet the stratosphere is full of ozone. Sydney Chapman explained why in 1930: ultraviolet light splits O₂ and the atoms find other O₂ molecules. "Never spontaneous" means that heat alone cannot do it, not that nothing can.

Problem solved in full

Where limestone really turns, from the hot tables 4 steps
From 25 °C values the page puts limestone's crossover at 843.1 °C, and a real kiln needs about 898 °C. Robie and Hemingway (USGS, 1995) also tabulate each substance at high temperature. At 1100 K, (H − H₂₉₈)/T is 81.61 J/(mol·K) for CaCO₃, 36.97 for CaO and 35.35 for CO₂, and S° is 232.23, 102.88 and 274.51. At 1200 K the six numbers are 85.57, 38.38, 37.06 and 243.46, 107.58, 279.37. Find where ΔG really crosses zero.

Open this problem in the tool
1. Heat content first. The products gain 36.97 + 35.35 J/(mol·K) and the carbonate 81.61, so the reaction's enthalpy falls by 9.29 × 1100 J, about 10.2 kJ below its 25 °C value.
  
  $Δ_{r} H (1100) = 178.8 + 1100 \times \frac{36 . 97 + 35 . 35 - 81 . 61}{1000} = 178.8 - 10.219 = 168.581 kJ/mol$
2. The entropy of reaction at 1100 K comes straight from the three S° values, and it is 15 J/(mol·K) smaller than at 25 °C.
  
  $Δ_{r} S (1100) = 102.88 + 274.51 - 232.23 = 145.16 J/ (mol K)$
3. So at 1100 K the reaction still refuses, by 8.905 kJ/mol. Repeating both steps at 1200 K gives ΔG = −5.544 kJ/mol: it goes.
  
  $Δ_{r} G (1100) = 168.581 - 1100 \times 0.14516 = + 8.905 Δ_{r} G (1200) = - 5.544 kJ/mol$
4. ΔG is nearly straight over a hundred kelvin, so the crossing sits 8.905/(8.905 + 5.544) of the way from 1100 K to 1200 K.
  
  $T = 1100 + 100 \times \frac{8 . 905}{8 . 905 + 5 . 544} = 1161.6 K$
Answer

About 1161.6 K, or 888.5 °C. Of the 55 degrees between the page's 843.1 °C and a working kiln, 45.4 are the constant-ΔH-and-ΔS approximation: both fall as the solid heats, and ΔS falls a little faster. The last 9.5 degrees is about what calcite's own ±1.3 kJ/mol uncertainty in ΔfH° allows, so the tables and Smyth and Adams's 898 °C agree as well as the data can. The page's straight line is the right tool for asking which way a reaction goes, and a rough one for asking at exactly what temperature.

$T \approx 1161.6 K = 888.5^{\circ} C 888.5 - 843.1 = 45.4$

Learning path

Why a reaction goes

References (3)

Every limestone figure on the page, and the high-temperature rows the worked problem uses: calcite, lime and carbon dioxide from one compilation, so that the crossover is not moved by mixing tables: R. A. Robie and B. S. Hemingway, Thermodynamic Properties of Minerals and Related Substances at 298.15 K and 1 Bar (10^5 Pascals) Pressure and at Higher Temperatures. U.S. Geological Survey Bulletin 2131, 1995. DOI 10.3133/b2131
The 898 °C at which calcite's CO2 pressure reaches one atmosphere, which insight block 1 and the worked problem measure the page against: F. H. Smyth and L. H. Adams, "The System, Calcium Oxide-Carbon Dioxide." Journal of the American Chemical Society 45(5), 1923, pp. 1167-1184. DOI 10.1021/ja01658a009
Ammonia, nitrogen, hydrogen and oxygen are CODATA key values and ozone is NIST-JANAF, each read from its WebBook page (this is ammonia's): P. J. Linstrom and W. G. Mallard (eds.), NIST Chemistry WebBook, NIST Standard Reference Database Number 69. National Institute of Standards and Technology, Gaithersburg MD. Ammonia, gas-phase thermochemistry.

By the 42Math team · Reviewed against published sources. Corrections welcome.